Division of Labor in Ant Colonies

Alex Cornejo, Anna Dornhaus,Nancy Lynch and Radhika Nagpal

October 13, 2014

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What is division of labor?

I Division of labor is the process by whichindividual ants in a colony decide which task toperform to ensure the survival of the colony.

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What is division of labor?

I Division of labor is the process by whichindividual ants in a colony decide which task toperform to ensure the survival of the colony.

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What is division of labor?

I Division of labor is the process by whichindividual ants in a colony decide which task toperform to ensure the survival of the colony.

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What is division of labor?

I Division of labor is the process by whichindividual ants in a colony decide which task toperform to ensure the survival of the colony.

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What is division of labor? (cont.)

I Much of the existing work focuses on exploringthe correlation between temporal andmorphological variations and worker behavior.

3/19

What is division of labor? (cont.)

I Much of the existing work focuses on exploringthe correlation between temporal andmorphological variations and worker behavior.

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What is division of labor? (cont.)

I Much of the existing work focuses on exploringthe correlation between temporal andmorphological variations and worker behavior.

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Contributions

I A very general mathematical formulation for thephenomenon of division of labor in ant colonies.

I A distributed algorithm that quickly reaches anear optimal task allocation and imposing onlyminimal assumptions on the capabilities ofindividual ants.

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Contributions

I A very general mathematical formulation for thephenomenon of division of labor in ant colonies.

I A distributed algorithm that quickly reaches anear optimal task allocation and imposing onlyminimal assumptions on the capabilities ofindividual ants.

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A model for division of labor

I A set A of ants.

I A set T of tasks.I For task τ ∈ T , ant a ∈ A and time t ∈ R≥0:

I d(τ, t) is the total energy demand for task τat time t.

I e(τ, a, t) is the energy that can be suppliedby ant a if engaged at task τ at time t.

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A model for division of labor

I A set A of ants.

I A set T of tasks.I For task τ ∈ T , ant a ∈ A and time t ∈ R≥0:

I d(τ, t) is the total energy demand for task τat time t.

I e(τ, a, t) is the energy that can be suppliedby ant a if engaged at task τ at time t.

5/19

A model for division of labor

I A set A of ants.

I A set T of tasks.

I For task τ ∈ T , ant a ∈ A and time t ∈ R≥0:

I d(τ, t) is the total energy demand for task τat time t.

I e(τ, a, t) is the energy that can be suppliedby ant a if engaged at task τ at time t.

5/19

A model for division of labor

I A set A of ants.

I A set T of tasks.I For task τ ∈ T , ant a ∈ A and time t ∈ R≥0:

I d(τ, t) is the total energy demand for task τat time t.

I e(τ, a, t) is the energy that can be suppliedby ant a if engaged at task τ at time t.

5/19

A model for division of labor

I A set A of ants.

I A set T of tasks.I For task τ ∈ T , ant a ∈ A and time t ∈ R≥0:

I d(τ, t) is the total energy demand for task τat time t.

I e(τ, a, t) is the energy that can be suppliedby ant a if engaged at task τ at time t.

5/19

A model for division of labor

I A set A of ants.

I A set T of tasks.I For task τ ∈ T , ant a ∈ A and time t ∈ R≥0:

I d(τ, t) is the total energy demand for task τat time t.

I e(τ, a, t) is the energy that can be suppliedby ant a if engaged at task τ at time t.

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Task Assignment

I A task assignment is a functiony : A× R≥0 → T ∪ {⊥}.

I Given y we define Y (τ, t) as the set of antsassigned to task τ at time t, and I(t) as the setof ants assigned to no task at time t.

Y (τ, t) = {a ∈ A : y(a, t) = τ}I(t) = {a ∈ A : y(a, t) =⊥}

I Observe that by definition A = Y (τ, t) ∪ I(t).

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Task Assignment

I A task assignment is a functiony : A× R≥0 → T ∪ {⊥}.

I Given y we define Y (τ, t) as the set of antsassigned to task τ at time t, and I(t) as the setof ants assigned to no task at time t.

Y (τ, t) = {a ∈ A : y(a, t) = τ}I(t) = {a ∈ A : y(a, t) =⊥}

I Observe that by definition A = Y (τ, t) ∪ I(t).

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Task Assignment

I A task assignment is a functiony : A× R≥0 → T ∪ {⊥}.

I Given y we define Y (τ, t) as the set of antsassigned to task τ at time t, and I(t) as the setof ants assigned to no task at time t.

Y (τ, t) = {a ∈ A : y(a, t) = τ}I(t) = {a ∈ A : y(a, t) =⊥}

I Observe that by definition A = Y (τ, t) ∪ I(t).

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Task Assignment

I A task assignment is a functiony : A× R≥0 → T ∪ {⊥}.

I Given y we define Y (τ, t) as the set of antsassigned to task τ at time t, and I(t) as the setof ants assigned to no task at time t.

Y (τ, t) = {a ∈ A : y(a, t) = τ}I(t) = {a ∈ A : y(a, t) =⊥}

I Observe that by definition A = Y (τ, t) ∪ I(t).

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What is a good task allocation?

I w(τ, t) =∑

a∈Y (τ,t) e(τ, a, t) is the energysupplied to task τ at time t.

I q(τ, t) = d(τ, t)− w(τ, t), if negative representsa surplus of energy at task τ , if positiverepresents a deficit of energy at task τ , and ifzero then the task τ is in equilibrium.

I An optimal task assignment is one thatminimizes

∑τ∈T q(τ, t)

2

7/19

What is a good task allocation?

I w(τ, t) =∑

a∈Y (τ,t) e(τ, a, t) is the energysupplied to task τ at time t.

I q(τ, t) = d(τ, t)− w(τ, t), if negative representsa surplus of energy at task τ , if positiverepresents a deficit of energy at task τ , and ifzero then the task τ is in equilibrium.

I An optimal task assignment is one thatminimizes

∑τ∈T q(τ, t)

2

7/19

What is a good task allocation?

I w(τ, t) =∑

a∈Y (τ,t) e(τ, a, t) is the energysupplied to task τ at time t.

I q(τ, t) = d(τ, t)− w(τ, t), if negative representsa surplus of energy at task τ , if positiverepresents a deficit of energy at task τ , and ifzero then the task τ is in equilibrium.

I An optimal task assignment is one thatminimizes

∑τ∈T q(τ, t)

2

7/19

What is a good task allocation?

I w(τ, t) =∑

a∈Y (τ,t) e(τ, a, t) is the energysupplied to task τ at time t.

I q(τ, t) = d(τ, t)− w(τ, t), if negative representsa surplus of energy at task τ , if positiverepresents a deficit of energy at task τ , and ifzero then the task τ is in equilibrium.

I An optimal task assignment is one thatminimizes

∑τ∈T q(τ, t)

2

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Restricted System Model

I We assume e(τ, a, t) = c for all τ ∈ T , everyant a ∈ A, and every time t ∈ R≥0.

I We consider synchronous model where timeproceeds in lock-step rounds 1, 2, . . ., and forthe duration of each round i ∈ N each anta ∈ A works at task y(a, i).

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Restricted System Model

I We assume e(τ, a, t) = c for all τ ∈ T , everyant a ∈ A, and every time t ∈ R≥0.

I We consider synchronous model where timeproceeds in lock-step rounds 1, 2, . . ., and forthe duration of each round i ∈ N each anta ∈ A works at task y(a, i).

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Restricted System Model

I We assume e(τ, a, t) = c for all τ ∈ T , everyant a ∈ A, and every time t ∈ R≥0.

I We consider synchronous model where timeproceeds in lock-step rounds 1, 2, . . ., and forthe duration of each round i ∈ N each anta ∈ A works at task y(a, i).

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Capabilities of Individuals

I Recall that q(τ, i) = d(τ, i)− w(τ, i) is thedifference between the energy demand for taskτ at round i and the energy supplied for task τat round i.

I For each task τ ∈ T ants can sense only abinary feedback function f(τ, i) where:

f(τ, i) =

{+1 q(τ, i) ≥ 0,

−1 q(τ, i) < 0.

9/19

Capabilities of Individuals

I Recall that q(τ, i) = d(τ, i)− w(τ, i) is thedifference between the energy demand for taskτ at round i and the energy supplied for task τat round i.

I For each task τ ∈ T ants can sense only abinary feedback function f(τ, i) where:

f(τ, i) =

{+1 q(τ, i) ≥ 0,

−1 q(τ, i) < 0.

9/19

Capabilities of Individuals

I Recall that q(τ, i) = d(τ, i)− w(τ, i) is thedifference between the energy demand for taskτ at round i and the energy supplied for task τat round i.

I For each task τ ∈ T ants can sense only abinary feedback function f(τ, i) where:

f(τ, i) =

{+1 q(τ, i) ≥ 0,

−1 q(τ, i) < 0.

9/19

Algorithm Idea

I Intuition: Given sufficient memory, ants coulduse a randomized binary-search-like strategy,guided by the function f , to reach the optimaldivision of labor.

I Idea 1: Offload the burden of memory from theindividuals to the colony.

I Idea 2: Let ants use a response-thresholdstrategy to pick tasks, and allow them tospecialize.

10/19

Algorithm Idea

I Intuition: Given sufficient memory, ants coulduse a randomized binary-search-like strategy,guided by the function f , to reach the optimaldivision of labor.

I Idea 1: Offload the burden of memory from theindividuals to the colony.

I Idea 2: Let ants use a response-thresholdstrategy to pick tasks, and allow them tospecialize.

10/19

Algorithm Idea

I Intuition: Given sufficient memory, ants coulduse a randomized binary-search-like strategy,guided by the function f , to reach the optimaldivision of labor.

I Idea 1: Offload the burden of memory from theindividuals to the colony.

I Idea 2: Let ants use a response-thresholdstrategy to pick tasks, and allow them tospecialize.

10/19

Algorithm Idea

I Intuition: Given sufficient memory, ants coulduse a randomized binary-search-like strategy,guided by the function f , to reach the optimaldivision of labor.

I Idea 1: Offload the burden of memory from theindividuals to the colony.

I Idea 2: Let ants use a response-thresholdstrategy to pick tasks, and allow them tospecialize.

10/19

Algorithm Details

I Each ant stores a task currentTask ∈ T ∪ {⊥}and a potential table %[τ ].

I Initially ants start in the Resting state withcurrentTask =⊥ and a potential of zero forevery task ∀τ ∈ T, %[τ ] = 0.

I Ants can be in one of the five states Resting,FirstReserve, SecondReserve,TempWorker and CoreWorker.

11/19

Algorithm Details

I Each ant stores a task currentTask ∈ T ∪ {⊥}and a potential table %[τ ].

I Initially ants start in the Resting state withcurrentTask =⊥ and a potential of zero forevery task ∀τ ∈ T, %[τ ] = 0.

I Ants can be in one of the five states Resting,FirstReserve, SecondReserve,TempWorker and CoreWorker.

11/19

Algorithm Details

I Each ant stores a task currentTask ∈ T ∪ {⊥}and a potential table %[τ ].

I Initially ants start in the Resting state withcurrentTask =⊥ and a potential of zero forevery task ∀τ ∈ T, %[τ ] = 0.

I Ants can be in one of the five states Resting,FirstReserve, SecondReserve,TempWorker and CoreWorker.

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Simplified State Machine

FirstReserve SecondReserve TempWorker CoreWorkerResting

threshold

+1 prob 12

+1 prob 12

+1 +1

−1

−1

−1

−1

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Resting State

∀τ ∈ T, %[τ ]←

{0 if f(τ, i) < 0

min(%[τ ] + 1, 3) if f(τ, i) > 0

candidate ← {τ ∈ T | %[τ ] = 3}if candidate 6= ∅ then

with probability 12do

∀τ ∈ T, %[τ ]← 0currentTask ← random candidate taskstate ← TempWorker

end withend if

13/19

Resting State

∀τ ∈ T, %[τ ]←

{0 if f(τ, i) < 0

min(%[τ ] + 1, 3) if f(τ, i) > 0

candidate ← {τ ∈ T | %[τ ] = 3}

if candidate 6= ∅ thenwith probability 1

2do

∀τ ∈ T, %[τ ]← 0currentTask ← random candidate taskstate ← TempWorker

end withend if

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Resting State

∀τ ∈ T, %[τ ]←

{0 if f(τ, i) < 0

min(%[τ ] + 1, 3) if f(τ, i) > 0

candidate ← {τ ∈ T | %[τ ] = 3}if candidate 6= ∅ then

with probability 12do

∀τ ∈ T, %[τ ]← 0currentTask ← random candidate taskstate ← TempWorker

end withend if

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Resting State

∀τ ∈ T, %[τ ]←

{0 if f(τ, i) < 0

min(%[τ ] + 1, 3) if f(τ, i) > 0

candidate ← {τ ∈ T | %[τ ] = 3}if candidate 6= ∅ then

with probability 12do

∀τ ∈ T, %[τ ]← 0currentTask ← random candidate taskstate ← TempWorker

end withend if

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Working States

case state = TempWorker

if f(currentTask, i) < 0 thenstate ← FirstReserve

elsestate ← CoreWorker

end ifcase state = CoreWorker

if f(currentTask, i) < 0 thenstate ← TempWorker

end ifend case

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Working States

case state = TempWorkerif f(currentTask, i) < 0 then

state ← FirstReserve

elsestate ← CoreWorker

end ifcase state = CoreWorker

if f(currentTask, i) < 0 thenstate ← TempWorker

end ifend case

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Working States

case state = TempWorkerif f(currentTask, i) < 0 then

state ← FirstReserveelse

state ← CoreWorkerend if

case state = CoreWorkerif f(currentTask, i) < 0 then

state ← TempWorkerend if

end case

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Working States

case state = TempWorkerif f(currentTask, i) < 0 then

state ← FirstReserveelse

state ← CoreWorkerend if

case state = CoreWorker

if f(currentTask, i) < 0 thenstate ← TempWorker

end ifend case

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Working States

case state = TempWorkerif f(currentTask, i) < 0 then

state ← FirstReserveelse

state ← CoreWorkerend if

case state = CoreWorkerif f(currentTask, i) < 0 then

state ← TempWorkerend if

end case

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Reserve Statescase state = FirstReserve

if f(currentTask, i) < 0 thenstate ← Resting

elsewith probability 1

2do

state ← TempWorkerotherwise

state ← SecondReserveend with

end ifcase state = SecondReserve

if f(currentTask, i) < 0 thenstate ← Resting

elsestate ← TempWorker

end ifend case

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Reserve Statescase state = FirstReserve

if f(currentTask, i) < 0 thenstate ← Resting

else

with probability 12do

state ← TempWorkerotherwise

state ← SecondReserveend with

end ifcase state = SecondReserve

if f(currentTask, i) < 0 thenstate ← Resting

elsestate ← TempWorker

end ifend case

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Reserve Statescase state = FirstReserve

if f(currentTask, i) < 0 thenstate ← Resting

elsewith probability 1

2do

state ← TempWorkerotherwise

state ← SecondReserveend with

end if

case state = SecondReserveif f(currentTask, i) < 0 then

state ← Restingelse

state ← TempWorkerend if

end case

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Reserve Statescase state = FirstReserve

if f(currentTask, i) < 0 thenstate ← Resting

elsewith probability 1

2do

state ← TempWorkerotherwise

state ← SecondReserveend with

end ifcase state = SecondReserve

if f(currentTask, i) < 0 thenstate ← Resting

elsestate ← TempWorker

end ifend case

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Reserve Statescase state = FirstReserve

if f(currentTask, i) < 0 thenstate ← Resting

elsewith probability 1

2do

state ← TempWorkerotherwise

state ← SecondReserveend with

end ifcase state = SecondReserve

if f(currentTask, i) < 0 thenstate ← Resting

elsestate ← TempWorker

end ifend case

15/19

Reserve Statescase state = FirstReserve

if f(currentTask, i) < 0 thenstate ← Resting

elsewith probability 1

2do

state ← TempWorkerotherwise

state ← SecondReserveend with

end ifcase state = SecondReserve

if f(currentTask, i) < 0 thenstate ← Resting

elsestate ← TempWorker

end ifend case

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Analysis Outline

I Lemma 1: If during an interval tasks are nottoo oversatisfied, the probability there is a taskwith a deficit is exponentially small in theinterval length.

16/19

Analysis Outline

I Lemma 1: If during an interval tasks are nottoo oversatisfied, the probability there is a taskwith a deficit is exponentially small in theinterval length.

16/19

Analysis Outline

I Lemma 1: If during an interval tasks are nottoo oversatisfied, the probability there is a taskwith a deficit is exponentially small in theinterval length.

time

ants

0

16/19

Analysis Outline

I Lemma 1: If during an interval tasks are nottoo oversatisfied, the probability there is a taskwith a deficit is exponentially small in theinterval length.

time

ants

0

16/19

Analysis Outline

I Lemma 1: If during an interval tasks are nottoo oversatisfied, the probability there is a taskwith a deficit is exponentially small in theinterval length.

time

ants

0

16/19

Analysis Outline

I Lemma 1: If during an interval tasks are nottoo oversatisfied, the probability there is a taskwith a deficit is exponentially small in theinterval length.

time

ants

0

16/19

Analysis Outline

I Lemma 1: If during an interval tasks are nottoo oversatisfied, the probability there is a taskwith a deficit is exponentially small in theinterval length.

time

ants

0

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Analysis Outline

I Lemma 2: Once a task transitions from havinga deficit to a surplus, it will keep oscillatingevery two or three rounds; moreover, withconstant probability the number of ants involvedin the oscillations is reduced by a constantfraction.

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Analysis Outline

I Lemma 2: Once a task transitions from havinga deficit to a surplus, it will keep oscillatingevery two or three rounds; moreover, withconstant probability the number of ants involvedin the oscillations is reduced by a constantfraction.

time

ants

0

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Analysis Outline

I Lemma 2: Once a task transitions from havinga deficit to a surplus, it will keep oscillatingevery two or three rounds; moreover, withconstant probability the number of ants involvedin the oscillations is reduced by a constantfraction.

time

ants

0

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Analysis Outline

I Lemma 2: Once a task transitions from havinga deficit to a surplus, it will keep oscillatingevery two or three rounds; moreover, withconstant probability the number of ants involvedin the oscillations is reduced by a constantfraction.

time

ants

0

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Analysis Outline

I Lemma 2: Once a task transitions from havinga deficit to a surplus, it will keep oscillatingevery two or three rounds; moreover, withconstant probability the number of ants involvedin the oscillations is reduced by a constantfraction.

time

ants

0

17/19

Analysis Outline

I Lemma 2: Once a task transitions from havinga deficit to a surplus, it will keep oscillatingevery two or three rounds; moreover, withconstant probability the number of ants involvedin the oscillations is reduced by a constantfraction.

time

ants

0

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Analysis Outline

I Theorem: After O(log n) rounds, with highprobability, ants reach an optimal taskallocation.

time

ants

0

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Questions?

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